3093: [Fdu校赛2012] A Famous Game
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 242 Solved: 129
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Description
Mr. B and Mr. M like to play with balls. They have many balls colored in blue and red. Firstly, Mr. B randomly picks up N balls out of them and put them into a bag. Mr. M knows that there are N+1 possible situations in which the number of red balls is ranged from 0 to N, and we assume the possibilities of the N+1 situations are the same. But Mr. M does not know which situation occurs. Secondly, Mr. M picks up P balls out of the bag and examines them. There are Q red balls and P-Q blue balls. The question is: if he picks up one more ball out of the bag, what is the possibility that this ball is red?
Input
Each test case contains only one line with three integers N, P and Q (2 <= N <= 100,000, 0 <= P <= N-1, 0 <= Q <= P).
Output
For each test case, display a single line containing the case number and the possibility of the next ball Mr. M picks out is red. The number should be rounded to four decimal places.
Sample Input
3 0 0
4 2 1
Sample Output
Case 1: 0.5000
Case 2: 0.5000
HINT
[Explanation]
For example as the sample test one, there are three balls in the bag. The possibilities of the four possible situations are all 0.25. If there are no red balls in the bag, the possibility of the next ball are red is 0. If there is one red ball in the bag, the possibility is 1/3. If there are two red balls, the possibility is 2/3. Finally if all balls are red, the possibility is 1. So the answer is 0*(1/4)+(1/3)*(1/4)+(2/3)*(1/4)+1*(1/4)=0.5.
Source
题意：已知有N个球，其中K个是蓝的，其余的是红的。K的范围是0到N，并且等概率。某人从其中随机取球（不放回），现在取出了P个，其中Q个是蓝色的，求下一个球是蓝色的概率。
今天学习的概率初等内容，然而蒟蒻的我只推出了$O(N)$的做法，式子经ZXR大佬化简后得到——$frac{q+1}{p+2}$，呵呵，怪不得时限给1s。
1 #include <cstdio> 2 3 signed main(void) 4 { 5 int p, q, c = 0; 6 7 while (scanf("%*d%d%d", &p, &q) != EOF) 8 printf("Case %d: %.4lf ", ++c, 1.0*(q+1)/(p+2)); 9 }
@Author: YouSiki